Lagrangian Equation of Motion for Planetary Orbit in a Single Plane 
Assume that a mass $m$ is in a gravitational orbit around a much larger mass $M$, as in the case of the earth revolving around the sun. Also, assume the motion is constrained to a single horizontal $xy$ plane.

I set up the Lagrangian equation of motion in polar coordinates as follows:
$\large L = T - V$
$\large L = \frac{1}{2}m(\dot{r}^2+(r\dot{\theta})^2)+\frac{GMm}{r}$
Euler-Langrange equation for polar coordinates:
$\large\frac{\partial{L}}{\partial{\theta}}-\frac{d}{dt}(\frac{\partial{L}}{\partial{\dot{\theta}}}) = 0$ $\rightarrow$
$\large\frac{d}{dt}(mr^2\dot{\theta}) = m(2r\dot{r}\dot{\theta}+r^2\ddot{\theta}) = 0$ $\rightarrow$
(assuming the trivial solution $r=0$ is not the answer, i.e. the orbiting has a non-zero potential)
$\large\mathbf{2\dot{r}\dot{\theta}+r\ddot{\theta} = 0}$
Here is where I am stuck.
Assuming that I performed the above steps correctly, how can I solve this differential equation in $r$ and $\theta$ to come up with some generalized equation of motion? I know that Kepler's Law dictates that a stable orbit must be an ellipse. But here is where my skill with differential equations and conical sections fails me.
 A: Let's start from
$$
L = \frac{1}{2}(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)
$$
There are two degrees of freedom: $r$ and $\theta$. Start with $\theta$

$\theta$

$$
\frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial \dot{\theta}} = 0 = \frac{{\rm d}}{{\rm d}t}(r^2\dot{\theta})~~~\Rightarrow~~~ r^2\dot{\theta} = l = {\rm const} \tag{1}
$$
which means that the number $l = r^2\dot{\theta}$ is a constant (angular momentum!).

$r$

$$
\frac{{\rm d}}{{\rm d}t}\frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} = 0 = \ddot{r} - r\dot{\theta}^2 + \frac{{\rm d}V}{{\rm d}r} \tag{2}
$$
Use (1) to write
$$
\frac{{\rm d}}{{\rm d}t} = \frac{l}{r^2}\frac{{\rm d}}{{\rm d}\theta}
$$
and define the variable $u = 1/r$, if you replace both into (2) you will get
$$
\frac{{\rm d}^2u}{{\rm d}t^2} + u = \frac{1}{l^2 u^2}\frac{{\rm d}V(1/u)}{{\rm d}r} \tag{3}
$$
For a Keppler potential
$$
V(r) = -\frac{GM}{r} = -GMu
$$
Replace that in (3) and you get
$$
\frac{{\rm d}^2u}{{\rm d}t^2} + u = \frac{GM}{l^2}
$$
whose solution is
$$
u(\theta) = C\cos(\theta - \theta_0) + \frac{GM}{l^2} 
$$
Now define 
$$
e = \frac{Cl^2}{GM} ~~~\mbox{and}~~~ a = \frac{l^2}{GM(1 - e^2)}
$$
such that the equation above becomes
$$ \bbox[5px,border:2px solid blue]
{
r(\theta) = \frac{a(1 - e^2)}{1 + e\cos(\theta - \theta_0)}
}
\tag{4}
$$
And there you have it, the solution are conics 
